The altitudes $BB_1$ and $CC_1$ of an acute triangle $ABC$ intersect in point $H$. Let $B_2$ and $C_2$ be points on the segments $BH$ and $CH$, respectively, such that $BB_2=B_1H$ and $CC_2=C_1H$. The circumcircle of the triangle $B_2HC_2$ intersects the circumcircle of the triangle $ABC$ in points $D$ and $E$. Prove that the triangle $DEH$ is right-angled.
Problem
Source: Baltic Way 2018, Problem 12
Tags: geometry
PROF65
06.11.2018 13:58
let $I,J,K,H'$ be the midpoints of $BC, BB_1,CC_1$ and the foot of $H$ on $BC$ remark that $HH'IJK$ is cyclic and $B_2 ,C_2$are symmetrc in $J,K$ so applying an homothety with center $H$ and ratio 2 sends the circle $(HJK) $ to $(HB_2C_2)$ and $HIH'$ to $HDE$ so the result follows. RH HAS
Tintarn
06.11.2018 14:22
Let $A'$ be the point on the circumcircle opposite to $A$. We prove that $\angle HC_2A'=\angle HB_2A'=90^\circ$. This will prove that $A'$ is one of the points $D$ and $E$ and $HA'$ is the diameter.
By symmetry, it suffices to prove $\angle HB_2A'=90^\circ$.
Let $F$ be the foot of the perpendicular from $A'$ to $BB_1$. We will prove that $BF=B_1H$ which will clearly imply $F=B_2$ and hence the result.
Let $X$ be the point where $BB_1$ intersects the circumcircle for the second time. By symmetry we have $B_1X=BF$. Moreover it is well-known and easy to prove that $B_1X=B_1H$ (that is, if we reflect the orthocenter over one of the sides, we end up on the circumcircle). The claim follows.